Question

A chemical engineer studying the properties of fuels placed 1.670 g of a hydrocarbon in the bomb of a calorimeter and filled it with O2 gas. The bomb was immersed in 2.550 L of water and the reaction initiated. The water temperature rose from 20.00°C to 23.55°C. If the calorimeter (excluding the water) had a heat capacity of 403 J/K, what was the heat of reaction for combustion (qV) per gram of the fuel?

Answer 1

856,68 J/gram

Step-by-step explanation:

We know that the variaton of temperature of 1K is equal to 1°C

so if the ΔT is 3.55°C, it is 3.55K.

so

403 J - 1 K

x - 3.55K

X= 403 * 3.55 = 1430,65 J

1.670 g - 1430,65 J

1 g - y

y =
`(1430,65)/(1.670)` = 856,68 J

Answer 2

The heat of reaction for combustion per gram of the fuel is approximately 23023 J/g.

How to find heat of reaction?

Calculate the total heat absorbed by the water and calorimeter:

Temperature change (ΔT) = 23.55°C - 20.00°C

= 3.55°C

Heat capacity of water (
`c_w`) = 4.184 J/g°C

Heat absorbed by water (
`q_{water`) =
`m_(water) * c_w * \Delta T`

Mass of water (
`m_{water`) = 2.550 L × 1000 g/L

= 2550 g

`q_{water` = 2550 g × 4.184 J/g°C × 3.55°C

≈ 37145 J

Heat absorbed by calorimeter (
`q_{cal`) =
`c_(cal) * \Delta T`

Heat capacity of calorimeter (
`c_{cal`) = 403 J/K

`q_(cal) = 403 J/K * 3.55 K`

≈ 1420 J

`\text{Total heat absorbed} (q_(total)) = q_(water) + q_(cal)`

≈ 37145 J + 1420 J

≈ 38565 J

Calculate the heat of reaction per gram of fuel:

`\text{Heat of reaction per gram} ((q_V)/(g)) = \frac{q_(total)}{\text{mass of fuel}}`

`(q_V)/(g ) = (38565 J)/(1.670 g)`

≈ 23023 J/g

Therefore, the heat of reaction for combustion per gram of the fuel is approximately 23023 J/g.